Find Each Of The Following Functions And State Their Domains

Unveiling the secrets of function composition and domain determination unveils a deeper understanding of how mathematical relationships interact and the boundaries within which they operate.

Understanding Function Composition

Function composition is a process where one function is applied to the result of another. Imagine it as a mathematical assembly line: one function takes an input, processes it, and passes the output to another function for further processing.

Formal Definition: Given two functions, f(x) and g(x), the composition of f with g, denoted as f(g(x)), is defined as applying the function f to the result of applying the function g to x. In simpler terms, we first evaluate g(x), then use that result as the input for f(x).

Notation: The composition f(g(x)) is read as "f of g of x." It can also be written as (f ∘ g)(x), where the circle symbol (∘) represents composition.

Example: Let's say f(x) = x² and g(x) = x + 1. Then, f(g(x)) = f(x + 1) = (x + 1)². This means we first add 1 to x using g(x), and then square the result using f(x).

Determining the Domain of a Function

The domain of a function is the set of all possible input values (often x-values) for which the function produces a valid output. Think of it as the set of numbers that "work" in the function without causing any mathematical errors.

Common Restrictions on Domains: Several factors can limit the domain of a function:

• Division by Zero: A denominator cannot be zero. If a function involves a fraction, we must exclude any x-values that make the denominator equal to zero.
• Square Roots of Negative Numbers: In the realm of real numbers, we cannot take the square root (or any even root) of a negative number. Therefore, the expression under the radical must be greater than or equal to zero.
• Logarithms of Non-Positive Numbers: Logarithms are only defined for positive numbers. The argument of a logarithm (the expression inside the logarithm) must be strictly greater than zero.
• Other Restrictions: Some functions might have inherent restrictions based on their specific definitions, such as trigonometric functions (e.g., tan(x) is undefined at certain values).

How to Find the Domain:

1. Identify Potential Restrictions: Look for the presence of fractions, square roots, logarithms, or other potentially problematic operations within the function.

2. Set Up Inequalities/Equations: Formulate inequalities or equations based on the restrictions identified in step 1. For example:

   • If there's a denominator, set it not equal to zero.
   • If there's a square root, set the expression under the radical greater than or equal to zero.
   • If there's a logarithm, set the argument of the logarithm greater than zero.

3. Solve for x: Solve the inequalities or equations to find the values of x that must be excluded from the domain.

4. Express the Domain: Write the domain in interval notation or set notation. Interval notation uses parentheses and brackets to indicate whether endpoints are included or excluded. Set notation uses curly braces to define the set of allowed values.

Finding Functions and Their Domains: Examples

Now, let's dive into some examples where we'll find composite functions and determine their domains.

Example 1:

• f(x) = √(x - 2)
• g(x) = x²

a) Find f(g(x))

• f(g(x)) = f(x²) = √(x² - 2)

b) Find the domain of f(g(x))

• The expression under the square root must be greater than or equal to zero: x² - 2 ≥ 0
• x² ≥ 2
• Taking the square root of both sides: |x| ≥ √2
• This means x ≥ √2 or x ≤ -√2
• Domain of f(g(x)): (-∞, -√2] ∪ [√2, ∞)

c) Find g(f(x))

• g(f(x)) = g(√(x - 2)) = (√(x - 2))² = x - 2

d) Find the domain of g(f(x))

• The original function f(x) has a square root, so x - 2 ≥ 0
• x ≥ 2
• Domain of g(f(x)): [2, ∞) Notice that even though the simplified form of g(f(x)) is x-2, we must consider the domain of the original composite function, which is restricted by the square root in f(x).

Example 2:

• f(x) = 1/x
• g(x) = x + 3

a) Find f(g(x))

• f(g(x)) = f(x + 3) = 1/(x + 3)

b) Find the domain of f(g(x))

• The denominator cannot be zero: x + 3 ≠ 0
• x ≠ -3
• Domain of f(g(x)): (-∞, -3) ∪ (-3, ∞)

c) Find g(f(x))

• g(f(x)) = g(1/x) = (1/x) + 3

d) Find the domain of g(f(x))

• The denominator cannot be zero: x ≠ 0
• Domain of g(f(x)): (-∞, 0) ∪ (0, ∞)

Example 3:

• f(x) = √(x + 4)
• g(x) = x² - 4

a) Find f(g(x))

• f(g(x)) = f(x² - 4) = √((x² - 4) + 4) = √(x²) = |x|

b) Find the domain of f(g(x))

• The expression under the square root in the original composite function must be greater than or equal to zero: (x² - 4) + 4 ≥ 0
• x² ≥ 0
• This is true for all real numbers.
• Domain of f(g(x)): (-∞, ∞)

c) Find g(f(x))

• g(f(x)) = g(√(x + 4)) = (√(x + 4))² - 4 = (x + 4) - 4 = x

d) Find the domain of g(f(x))

• The square root in f(x) restricts the domain: x + 4 ≥ 0
• x ≥ -4
• Domain of g(f(x)): [-4, ∞)

Example 4:

• f(x) = ln(x)
• g(x) = x² + 1

a) Find f(g(x))

• f(g(x)) = f(x² + 1) = ln(x² + 1)

b) Find the domain of f(g(x))

• The argument of the logarithm must be greater than zero: x² + 1 > 0
• Since x² is always non-negative, x² + 1 is always greater than zero for all real numbers.
• Domain of f(g(x)): (-∞, ∞)

c) Find g(f(x))

• g(f(x)) = g(ln(x)) = (ln(x))² + 1

d) Find the domain of g(f(x))

• The argument of the logarithm must be greater than zero: x > 0
• Domain of g(f(x)): (0, ∞)

Example 5:

• f(x) = 1/(x - 1)
• g(x) = √(x)

a) Find f(g(x))

• f(g(x)) = f(√(x)) = 1/(√(x) - 1)

b) Find the domain of f(g(x))

• We have two restrictions: the expression under the square root must be greater than or equal to zero, and the denominator cannot be zero.
• x ≥ 0
• √(x) - 1 ≠ 0 => √(x) ≠ 1 => x ≠ 1
• Combining these restrictions: x ≥ 0 and x ≠ 1
• Domain of f(g(x)): [0, 1) ∪ (1, ∞)

c) Find g(f(x))

• g(f(x)) = g(1/(x - 1)) = √(1/(x - 1))

d) Find the domain of g(f(x))

• We have two restrictions: the denominator cannot be zero, and the expression under the square root must be greater than or equal to zero. However, since the expression under the square root is a fraction, we need the entire fraction to be greater than or equal to zero.
• x - 1 ≠ 0 => x ≠ 1
• 1/(x - 1) ≥ 0. For a fraction to be positive, either both the numerator and denominator are positive, or both are negative. Since the numerator is 1 (positive), the denominator must also be positive.
• x - 1 > 0 => x > 1
• Domain of g(f(x)): (1, ∞)

Example 6:

• f(x) = 3x + 2
• g(x) = (x - 2)/3

a) Find f(g(x))

• f(g(x)) = f((x - 2)/3) = 3((x - 2)/3) + 2 = (x - 2) + 2 = x

b) Find the domain of f(g(x))

• There are no restrictions in either f(x) or g(x).
• Domain of f(g(x)): (-∞, ∞)

c) Find g(f(x))

• g(f(x)) = g(3x + 2) = ((3x + 2) - 2)/3 = (3x)/3 = x

d) Find the domain of g(f(x))

• There are no restrictions in either f(x) or g(x).
• Domain of g(f(x)): (-∞, ∞)

Notice that in this case, f(g(x)) = g(f(x)) = x. This means f(x) and g(x) are inverse functions of each other.

Example 7:

• f(x) = √(-x)
• g(x) = x + 1

a) Find f(g(x))

• f(g(x)) = f(x + 1) = √(-(x + 1)) = √(-x - 1)

b) Find the domain of f(g(x))

• The expression under the square root must be greater than or equal to zero: -x - 1 ≥ 0
• -x ≥ 1
• x ≤ -1
• Domain of f(g(x)): (-∞, -1]

c) Find g(f(x))

• g(f(x)) = g(√(-x)) = √(−x) + 1

d) Find the domain of g(f(x))

• The expression under the square root must be greater than or equal to zero: -x ≥ 0
• x ≤ 0
• Domain of g(f(x)): (-∞, 0]

Example 8:

• f(x) = |x| / x
• g(x) = x² - 1

a) Find f(g(x))

• f(g(x)) = f(x² - 1) = |x² - 1| / (x² - 1)

b) Find the domain of f(g(x))

• The denominator cannot be zero: x² - 1 ≠ 0
• x² ≠ 1
• x ≠ 1 and x ≠ -1
• Domain of f(g(x)): (-∞, -1) ∪ (-1, 1) ∪ (1, ∞)

c) Find g(f(x))

• g(f(x)) = g(|x|/x) = (|x|/x)² - 1 = (x²/x²) - 1 = 1 - 1 = 0, for x ≠ 0

d) Find the domain of g(f(x))

• The original function f(x) has a restriction: x ≠ 0
• Domain of g(f(x)): (-∞, 0) ∪ (0, ∞)

Example 9:

• f(x) = e^x (the exponential function)
• g(x) = x - 5

a) Find f(g(x))

• f(g(x)) = f(x - 5) = e^(x - 5)

b) Find the domain of f(g(x))

• Exponential functions are defined for all real numbers.
• Domain of f(g(x)): (-∞, ∞)

c) Find g(f(x))

• g(f(x)) = g(e^x) = e^x - 5

d) Find the domain of g(f(x))

• Exponential functions are defined for all real numbers.
• Domain of g(f(x)): (-∞, ∞)

Example 10:

• f(x) = arcsin(x) (the inverse sine function)
• g(x) = 2x

a) Find f(g(x))

• f(g(x)) = f(2x) = arcsin(2x)

b) Find the domain of f(g(x))

• The domain of arcsin(x) is [-1, 1]. Therefore, we must have:
• -1 ≤ 2x ≤ 1
• -1/2 ≤ x ≤ 1/2
• Domain of f(g(x)): [-1/2, 1/2]

c) Find g(f(x))

• g(f(x)) = g(arcsin(x)) = 2arcsin(x)*

d) Find the domain of g(f(x))

• The domain of arcsin(x) is [-1, 1].
• Domain of g(f(x)): [-1, 1]

Key Takeaways

• Function composition involves applying one function to the result of another. The order of composition matters (f(g(x)) is generally not the same as g(f(x))).
• The domain of a function is the set of all possible input values for which the function produces a valid output.
• Common domain restrictions arise from division by zero, square roots of negative numbers, and logarithms of non-positive numbers.
• When finding the domain of a composite function, consider the domain restrictions of both the inner and outer functions. The domain of the composite function is the set of all x values that are in the domain of the inner function g(x), and for which g(x) is in the domain of the outer function f(x). Even if the simplified expression of the composite function appears to have a larger domain, the original domain restrictions still apply.
• Always express the domain using interval notation or set notation.

Understanding function composition and domain determination is crucial for a solid foundation in mathematics, particularly in calculus and analysis. By mastering these concepts, you'll gain a deeper appreciation for the intricate relationships between functions and the boundaries within which they operate.